A 25.0 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution. A buffer region was found around a pH of 3.5. The unknown compound is

The unknown compound is a weak acid.

Explanation:

Given that :

a 25 mL sample of a solution of an unknown compound is titrated with a 0.115 M NaOH solution.

A buffer region was found around a pH of 3.5. We know that a pH of 3.5 is a weak acid. So, it is likely to be an organic acid

Let assume the solution of the unknown sample to be CH₃COOH

Now :

25 mL of CH₃COOH reacted with 0.115 M of NaOH

The equation for the reaction will be :

CH₃COOH    +    NaOH > CH₃COONa + H₂O

at x mole of      0.115y M of

CH₃COOH       NaOH is present

If NaOH was added in excess;

CH₃COOH    +    NaOH > CH₃COONa , NaOH will be lost then CH₃COOH    and CH₃COONa will be present

Therefore;

At equilibrium : Only CH₃COONa will be present but if it is above equilibrium NaOH will be present   because the pH will increase due to the presence of the strong base

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the answer is c. nacl